I'm working on modernizing Rosetta Code's infrastructure. Starting with communications. Please accept this time-limited open invite to RC's Slack.. --Michael Mol (talk) 20:59, 30 May 2020 (UTC)

Sum and product puzzle

Sum and product puzzle
You are encouraged to solve this task according to the task description, using any language you may know.

Solve the "Impossible Puzzle":

X and Y are two different whole numbers greater than 1. Their sum is no greater than 100, and Y is greater than X. S and P are two mathematicians (and consequently perfect logicians); S knows the sum X+Y and P knows the product X*Y. Both S and P know all the information in this paragraph.

The following conversation occurs:

• S says "P does not know X and Y."
• P says "Now I know X and Y."
• S says "Now I also know X and Y!"

What are X and Y?

Guidance

It can be hard to wrap one's head around what the three lines of dialog between S (the "sum guy") and P (the "product guy") convey about the values of X and Y.
So for your convenience, here's a break-down:

Quote Implied fact
1) S says "P does not know X and Y." For every possible sum decomposition of the number X+Y, the product has in turn more than one product decomposition.
2) P says "Now I know X and Y." The number X*Y has only one product decomposition for which fact 1 is true.
3) S says "Now I also know X and Y." The number X+Y has only one sum decomposition for which fact 2 is true.

Terminology:

• "sum decomposition" of a number = Any pair of positive integers (A, B) so that A+B equals the number. Here, with the additional constraint 2 ≤ A < B.
• "product decomposition" of a number = Any pair of positive integers (A, B) so that A*B equals the number. Here, with the additional constraint 2 ≤ A < B.

Your program can solve the puzzle by considering all possible pairs (X, Y) in the range 2 ≤ X < Y ≤ 98, and then successively eliminating candidates based on the three facts. It turns out only one solution remains!
See the Python example for an implementation that uses this approach with a few optimizations.

11l

Translation of: Python
`F counter(arr)   DefaultDict[Int, Int] d   L(a) arr      d[a]++   R d F decompose_sum(s)   R (2 .< Int(s / 2 + 1)).map(a -> (a, @s - a)) Set[(Int, Int)] all_pairs_setL(a) 2..99   L(b) a + 1 .< 100      I a + b < 100         all_pairs_set.add((a, b))V all_pairs = Array(all_pairs_set) V product_counts = counter(all_pairs.map((c, d) -> c * d))V unique_products = Set(all_pairs.filter((a, b) -> :product_counts[a * b] == 1))V s_pairs = all_pairs.filter((a, b) -> all(decompose_sum(a + b).map((x, y) -> (x, y) !C :unique_products))) product_counts = counter(s_pairs.map((c, d) -> c * d))V p_pairs = s_pairs.filter((a, b) -> :product_counts[a * b] == 1) V sum_counts = counter(p_pairs.map((c, d) -> c + d))V final_pairs = p_pairs.filter((a, b) -> :sum_counts[a + b] == 1) print(final_pairs)`
Output:
```[(4, 13)]
```

AWK

` # syntax: GAWK -f SUM_AND_PRODUCT_PUZZLE.AWKBEGIN {    for (s=2; s<=100; s++) {      if ((a=satisfies_statement3(s)) != 0) {        printf("%d (%d+%d)\n",s,a,s-a)      }    }    exit(0)}function satisfies_statement1(s,  a) { # S says: P does not know the two numbers.# Given s, for all pairs (a,b), a+b=s, 2 <= a,b <= 99, true if at least one of a or b is composite    for (a=2; a<=int(s/2); a++) {      if (is_prime(a) && is_prime(s-a)) {        return(0)      }    }    return(1)}function satisfies_statement2(p,  i,j,winner) { # P says: Now I know the two numbers.# Given p, for all pairs (a,b), a*b=p, 2 <= a,b <= 99, true if exactly one pair satisfies statement 1    for (i=2; i<=int(sqrt(p)); i++) {      if (p % i == 0) {        j = int(p/i)        if (!(2 <= j && j <= 99)) { # in range          continue        }        if (satisfies_statement1(i+j)) {          if (winner) {            return(0)          }          winner = 1        }      }    }    return(winner)}function satisfies_statement3(s,  a,b,winner) { # S says: Now I know the two numbers.# Given s, for all pairs (a,b), a+b=s, 2 <= a,b <= 99, true if exactly one pair satisfies statements 1 and 2    if (!satisfies_statement1(s)) {      return(0)    }    for (a=2; a<=int(s/2); a++) {      b = s - a      if (satisfies_statement2(a*b)) {        if (winner) {          return(0)        }        winner = a      }    }    return(winner)}function is_prime(x,  i) {    if (x <= 1) {      return(0)    }    for (i=2; i<=int(sqrt(x)); i++) {      if (x % i == 0) {        return(0)      }    }    return(1)} `

Output:

```17 (4+13)
```

C

Translation of: C#
`#include <stdbool.h>#include <stdio.h>#include <stdlib.h> typedef struct node_t {    int x, y;    struct node_t *prev, *next;} node; node *new_node(int x, int y) {    node *n = malloc(sizeof(node));    n->x = x;    n->y = y;    n->next = NULL;    n->prev = NULL;    return n;} void free_node(node **n) {    if (n == NULL) {        return;    }     (*n)->prev = NULL;    (*n)->next = NULL;     free(*n);     *n = NULL;} typedef struct list_t {    node *head;    node *tail;} list; list make_list() {    list lst = { NULL, NULL };    return lst;} void append_node(list *const lst, int x, int y) {    if (lst == NULL) {        return;    }     node *n = new_node(x, y);     if (lst->head == NULL) {        lst->head = n;        lst->tail = n;    } else {        n->prev = lst->tail;        lst->tail->next = n;        lst->tail = n;    }} void remove_node(list *const lst, const node *const n) {    if (lst == NULL || n == NULL) {        return;    }     if (n->prev != NULL) {        n->prev->next = n->next;        if (n->next != NULL) {            n->next->prev = n->prev;        } else {            lst->tail = n->prev;        }    } else {        if (n->next != NULL) {            n->next->prev = NULL;            lst->head = n->next;        }    }     free_node(&n);} void free_list(list *const lst) {    node *ptr;     if (lst == NULL) {        return;    }    ptr = lst->head;     while (ptr != NULL) {        node *nxt = ptr->next;        free_node(&ptr);        ptr = nxt;    }     lst->head = NULL;    lst->tail = NULL;} void print_list(const list *lst) {    node *it;     if (lst == NULL) {        return;    }     for (it = lst->head; it != NULL; it = it->next) {        int sum = it->x + it->y;        int prod = it->x * it->y;        printf("[%d, %d] S=%d P=%d\n", it->x, it->y, sum, prod);    }} void print_count(const list *const lst) {    node *it;    int c = 0;     if (lst == NULL) {        return;    }     for (it = lst->head; it != NULL; it = it->next) {        c++;    }     if (c == 0) {        printf("no candidates\n");    } else    if (c == 1) {        printf("one candidate\n");    } else {        printf("%d candidates\n", c);    }} void setup(list *const lst) {    int x, y;     if (lst == NULL) {        return;    }     // numbers must be greater than 1    for (x = 2; x <= 98; x++) {        // numbers must be unique, and sum no more than 100        for (y = x + 1; y <= 98; y++) {            if (x + y <= 100) {                append_node(lst, x, y);            }        }    }} void remove_by_sum(list *const lst, const int sum) {    node *it;     if (lst == NULL) {        return;    }     it = lst->head;    while (it != NULL) {        int s = it->x + it->y;         if (s == sum) {            remove_node(lst, it);            it = lst->head;        } else {            it = it->next;        }    }} void remove_by_prod(list *const lst, const int prod) {    node *it;     if (lst == NULL) {        return;    }     it = lst->head;    while (it != NULL) {        int p = it->x * it->y;         if (p == prod) {            remove_node(lst, it);            it = lst->head;        } else {            it = it->next;        }    }} void statement1(list *const lst) {    short *unique = calloc(100000, sizeof(short));    node *it, *nxt;     for (it = lst->head; it != NULL; it = it->next) {        int prod = it->x * it->y;        unique[prod]++;    }     it = lst->head;    while (it != NULL) {        int prod = it->x * it->y;        nxt = it->next;        if (unique[prod] == 1) {            remove_by_sum(lst, it->x + it->y);            it = lst->head;        } else {            it = nxt;        }    }     free(unique);} void statement2(list *const candidates) {    short *unique = calloc(100000, sizeof(short));    node *it, *nxt;     for (it = candidates->head; it != NULL; it = it->next) {        int prod = it->x * it->y;        unique[prod]++;    }     it = candidates->head;    while (it != NULL) {        int prod = it->x * it->y;        nxt = it->next;        if (unique[prod] > 1) {            remove_by_prod(candidates, prod);            it = candidates->head;        } else {            it = nxt;        }    }     free(unique);} void statement3(list *const candidates) {    short *unique = calloc(100, sizeof(short));    node *it, *nxt;     for (it = candidates->head; it != NULL; it = it->next) {        int sum = it->x + it->y;        unique[sum]++;    }     it = candidates->head;    while (it != NULL) {        int sum = it->x + it->y;        nxt = it->next;        if (unique[sum] > 1) {            remove_by_sum(candidates, sum);            it = candidates->head;        } else {            it = nxt;        }    }     free(unique);} int main() {    list candidates = make_list();     setup(&candidates);    print_count(&candidates);     statement1(&candidates);    print_count(&candidates);     statement2(&candidates);    print_count(&candidates);     statement3(&candidates);    print_count(&candidates);     print_list(&candidates);     free_list(&candidates);    return 0;}`
Output:
```2352 candidates
145 candidates
86 candidates
one candidate
[4, 13] S=17 P=52```

C++

Translation of: C
`#include <algorithm>#include <iostream>#include <map>#include <vector> std::ostream &operator<<(std::ostream &os, std::vector<std::pair<int, int>> &v) {    for (auto &p : v) {        auto sum = p.first + p.second;        auto prod = p.first * p.second;        os << '[' << p.first << ", " << p.second << "] S=" << sum << " P=" << prod;    }    return os << '\n';} void print_count(const std::vector<std::pair<int, int>> &candidates) {    auto c = candidates.size();    if (c == 0) {        std::cout << "no candidates\n";    } else if (c == 1) {        std::cout << "one candidate\n";    } else {        std::cout << c << " candidates\n";    }} auto setup() {    std::vector<std::pair<int, int>> candidates;     // numbers must be greater than 1    for (int x = 2; x <= 98; x++) {        // numbers must be unique, and sum no more than 100        for (int y = x + 1; y <= 98; y++) {            if (x + y <= 100) {                candidates.push_back(std::make_pair(x, y));            }        }    }     return candidates;} void remove_by_sum(std::vector<std::pair<int, int>> &candidates, const int sum) {    candidates.erase(std::remove_if(        candidates.begin(), candidates.end(),        [sum](const std::pair<int, int> &pair) {            auto s = pair.first + pair.second;            return s == sum;        }    ), candidates.end());} void remove_by_prod(std::vector<std::pair<int, int>> &candidates, const int prod) {    candidates.erase(std::remove_if(        candidates.begin(), candidates.end(),        [prod](const std::pair<int, int> &pair) {            auto p = pair.first * pair.second;            return p == prod;        }    ), candidates.end());} void statement1(std::vector<std::pair<int, int>> &candidates) {    std::map<int, int> uniqueMap;     std::for_each(        candidates.cbegin(), candidates.cend(),        [&uniqueMap](const std::pair<int, int> &pair) {            auto prod = pair.first * pair.second;            uniqueMap[prod]++;        }    );     bool loop;    do {        loop = false;        for (auto &pair : candidates) {            auto prod = pair.first * pair.second;            if (uniqueMap[prod] == 1) {                auto sum = pair.first + pair.second;                remove_by_sum(candidates, sum);                 loop = true;                break;            }        }    } while (loop);} void statement2(std::vector<std::pair<int, int>> &candidates) {    std::map<int, int> uniqueMap;     std::for_each(        candidates.cbegin(), candidates.cend(),        [&uniqueMap](const std::pair<int, int> &pair) {            auto prod = pair.first * pair.second;            uniqueMap[prod]++;        }    );     bool loop;    do {        loop = false;        for (auto &pair : candidates) {            auto prod = pair.first * pair.second;            if (uniqueMap[prod] > 1) {                remove_by_prod(candidates, prod);                 loop = true;                break;            }        }    } while (loop);} void statement3(std::vector<std::pair<int, int>> &candidates) {    std::map<int, int> uniqueMap;     std::for_each(        candidates.cbegin(), candidates.cend(),        [&uniqueMap](const std::pair<int, int> &pair) {            auto sum = pair.first + pair.second;            uniqueMap[sum]++;        }    );     bool loop;    do {        loop = false;        for (auto &pair : candidates) {            auto sum = pair.first + pair.second;            if (uniqueMap[sum] > 1) {                remove_by_sum(candidates, sum);                 loop = true;                break;            }        }    } while (loop);} int main() {    auto candidates = setup();    print_count(candidates);     statement1(candidates);    print_count(candidates);     statement2(candidates);    print_count(candidates);     statement3(candidates);    print_count(candidates);     std::cout << candidates;     return 0;}`
Output:
```2352 candidates
145 candidates
86 candidates
one candidate
[4, 13] S=17 P=52```

C#

`using System;using System.Linq;using System.Collections.Generic; public class Program{    public static void Main()    {        const int maxSum = 100;        var pairs = (            from X in 2.To(maxSum / 2 - 1)            from Y in (X + 1).To(maxSum - 2).TakeWhile(y => X + y <= maxSum)            select new { X, Y, S = X + Y, P = X * Y }            ).ToHashSet();         Console.WriteLine(pairs.Count);         var uniqueP = pairs.GroupBy(pair => pair.P).Where(g => g.Count() == 1).Select(g => g.Key).ToHashSet();         pairs.ExceptWith(pairs.GroupBy(pair => pair.S).Where(g => g.Any(pair => uniqueP.Contains(pair.P))).SelectMany(g => g));        Console.WriteLine(pairs.Count);         pairs.ExceptWith(pairs.GroupBy(pair => pair.P).Where(g => g.Count() > 1).SelectMany(g => g));        Console.WriteLine(pairs.Count);         pairs.ExceptWith(pairs.GroupBy(pair => pair.S).Where(g => g.Count() > 1).SelectMany(g => g));        Console.WriteLine(pairs.Count);         foreach (var pair in pairs) Console.WriteLine(pair);    }} public static class Extensions{    public static IEnumerable<int> To(this int start, int end) {        for (int i = start; i <= end; i++) yield return i;    }     public static HashSet<T> ToHashSet<T>(this IEnumerable<T> source) => new HashSet<T>(source);}`
Output:
```2352
145
86
1
{ X = 4, Y = 13, S = 17, P = 52 }
```

Common Lisp

Version 1

` ;;; Calculate all x's and their possible y's.(defparameter *x-possibleys*  (loop for x from 2 to 49     collect (cons x (loop for y from (- 100 x) downto (1+ x)			collect y)))  "For every x there are certain y's, with respect to the rules of the puzzle") (defun xys-operation (op x-possibleys)  "returns an alist of ((x possible-y) . (op x possible-y))"  (let ((x (car x-possibleys))	(ys (cdr x-possibleys)))    (mapcar #'(lambda (y) (cons (list x y) (funcall op x y))) ys))) (defun sp-numbers (op x-possibleys)  "generates all possible sums or products of the puzzle"  (loop for xys in x-possibleys       append (xys-operation op xys))) (defun group-sp (sp-numbers)  "sp: Sum or Product"  (loop for sp-number in (remove-duplicates sp-numbers :key #'cdr)     collect (cons (cdr sp-number)		   (mapcar #'car			   (remove-if-not			    #'(lambda (sp) (= sp (cdr sp-number)))			    sp-numbers			    :key #'cdr))))) (defun statement-1a (sum-groups)  "remove all sums with a single possible xy"  (remove-if   #'(lambda (xys) (= (list-length xys) 1))   sum-groups   :key #'cdr)) (defun statement-1b (x-possibleys)  "S says: P does not know X and Y."  (let ((multi-xy-sums (statement-1a (group-sp (sp-numbers #'+ x-possibleys))))	(products (group-sp (sp-numbers #'* x-possibleys))))    (flet ((sum-has-xy-which-leads-to-unique-prod (sum-xys)	     ;; is there any product with a single possible xy?	     (some #'(lambda (prod-xys) (= (list-length (cdr prod-xys)) 1))		   ;; all possible xys of the sum's (* x ys)		   (mapcar #'(lambda (xy) (assoc (apply #'* xy) products))			   (cdr sum-xys)))))      ;; remove sums with even one xy which leads to a unique product      (remove-if #'sum-has-xy-which-leads-to-unique-prod multi-xy-sums)))) (defun remaining-products (remaining-sums-xys)  "P's number is one of these"  (loop for sum-xys in remaining-sums-xys     append (loop for xy in (cdr sum-xys)	       collect (apply #'* xy)))) (defun statement-2 (remaining-sums-xys)  "P says: Now I know X and Y."  (let ((remaining-products (remaining-products remaining-sums-xys)))    (mapcar #'(lambda (a-sum-unit)		(cons (car a-sum-unit)		      (mapcar #'(lambda (xy)				  (list (count (apply #'* xy) remaining-products)					xy))			      (cdr a-sum-unit))))	    remaining-sums-xys))) (defun statement-3 (remaining-sums-with-their-products-occurrences-info)  "S says: Now I also know X and Y."  (remove-if   #'(lambda (sum-xys)       ;; remove those sums which have more than 1 product, that        ;; appear only once amongst all remaining products       (> (count 1 sum-xys :key #'car) 1))   remaining-sums-with-their-products-occurrences-info   :key #'cdr)) (defun solution (survivor-sum-and-its-xys)  "Now we know X and Y too :-D"  (let* ((sum (caar survivor-sum-and-its-xys))	 (xys (cdar survivor-sum-and-its-xys))	 (xy (second (find 1 xys :key #'car))))    (pairlis '(x y sum product)	     (list (first xy) (second xy) sum (apply #'* xy)))))  (solution (statement-3  (statement-2   (statement-1b *x-possibleys*)))) ;; => ((PRODUCT . 52) (SUM . 17) (Y . 13) (X . 4)) `

Version 2

` ;;; Algorithm of Rosetta code: ;;; All possible pairs(defparameter *all-possible-pairs*  (loop for i from 2 upto 100     append (loop for j from (1+ i) upto 100	       if (<= (+ i j) 100)	       collect (list i j)))) (defun oncep (item list)  (eql 1 (count item list))) ;;; Terminology(defun sum-decomp (n)  (loop for x from 2 below (/ n 2)     for y = (- n x)     collect (list x y))) (defun prod-decomp (n)  (loop for x from 2 below (sqrt n)     for y = (/ n x)        if (and (>= 100 (+ x y)) (zerop (rem n x)))     collect (list x y))) ;;; For every possible sum decomposition of the number X+Y, the product has in turn more than one product decomposition:(defun fact-1 (n)  "n = x + y"  (flet ((premise (pair)	       (> (list-length (prod-decomp (apply #'* pair))) 1)))	(every #'premise (sum-decomp n)))) ;;; The number X*Y has only one product decomposition for which fact 1 is true:(defun fact-2 (n)  "n = x * y"  (oncep t (mapcar (lambda (pair) (fact-1 (apply #'+ pair))) (prod-decomp n)))) ;;; The number X+Y has only one sum decomposition for which fact 2 is true:(defun fact-3 (n)  "n = x + y"  (oncep t (mapcar (lambda (pair) (fact-2 (apply #'* pair))) (sum-decomp n)))) (defun find-xy (all-possible-pairs)  (remove-if-not   #'(lambda (p) (fact-3 (apply #'+ p)))   (remove-if-not    #'(lambda (p) (fact-2 (apply #'* p)))    (remove-if-not     #'(lambda (p) (fact-1 (apply #'+ p)))     all-possible-pairs)))) (find-xy *all-possible-pairs*) ;; => ((4 13)) `

D

Translation of: Scala
`void main() {    import std.stdio, std.algorithm, std.range, std.typecons;     const s1 = cartesianProduct(iota(1, 101), iota(1, 101))               .filter!(p => 1 < p[0] && p[0] < p[1] && p[0] + p[1] < 100)               .array;     alias P = const Tuple!(int, int);    enum add   = (P p) => p[0] + p[1];    enum mul   = (P p) => p[0] * p[1];    enum sumEq = (P p) => s1.filter!(q => add(q) == add(p));    enum mulEq = (P p) => s1.filter!(q => mul(q) == mul(p));     const s2 = s1.filter!(p => sumEq(p).all!(q => mulEq(q).walkLength != 1)).array;    const s3 = s2.filter!(p => mulEq(p).setIntersection(s2).walkLength == 1).array;    s3.filter!(p => sumEq(p).setIntersection(s3).walkLength == 1).writeln;}`
Output:
`[const(Tuple!(int, int))(4, 13)]`

With an older version of the LDC2 compiler replace the `cartesianProduct` line with:

`     const s1 = iota(1, 101).map!(x => iota(1, 101).map!(y => tuple(x, y))).joiner `

The `.array` turn the lazy ranges into arrays. This is a necessary optimization because D lazy Ranges aren't memoized as Haskell lazy lists.

Run-time: about 0.43 seconds with dmd, 0.08 seconds with ldc2.

Elixir

Translation of: Ruby
`defmodule Puzzle do  def sum_and_product do    s1 = for x <- 2..49, y <- x+1..99, x+y<100, do: {x,y}    s2 = Enum.filter(s1, fn p ->      Enum.all?(sumEq(s1,p), fn q -> length(mulEq(s1,q)) != 1 end)    end)    s3 = Enum.filter(s2, fn p -> only1?(mulEq(s1,p), s2) end)    Enum.filter(s3, fn p -> only1?(sumEq(s1,p), s3) end) |> IO.inspect   end   defp add({x,y}), do: x + y   defp mul({x,y}), do: x * y   defp sumEq(s, p), do: Enum.filter(s, fn q -> add(p) == add(q) end)   defp mulEq(s, p), do: Enum.filter(s, fn q -> mul(p) == mul(q) end)   defp only1?(a, b) do    MapSet.size(MapSet.intersection(MapSet.new(a), MapSet.new(b))) == 1  endend Puzzle.sum_and_product`
Output:
```[{4, 13}]
```

Factor

A loose translation of D.

`USING: combinators.short-circuit fry kernel literals mathmath.ranges memoize prettyprint sequences sets tools.time ;IN: rosetta-code.sum-and-product CONSTANT: s1 \$[    2 100 [a,b] dup cartesian-product concat    [ first2 { [ < ] [ + 100 < ] } 2&& ] filter] : quot-eq ( pair quot -- seq )    [ s1 ] 2dip tuck '[ @ _ @ = ] filter ; inline MEMO: sum-eq ( pair -- seq ) [ first2 + ] quot-eq ;MEMO: mul-eq ( pair -- seq ) [ first2 * ] quot-eq ; : s2 ( -- seq )    s1 [ sum-eq [ mul-eq length 1 = not ] all? ] filter ; : only-1 ( seq quot -- newseq )    over '[ @ _ intersect length 1 = ] filter ; inline : sum-and-product ( -- )    [ s2 [ mul-eq ] [ sum-eq ] [ only-1 ] [email protected] . ] time ; MAIN: sum-and-product`
Output:
```{ { 4 13 } }
Running time: 0.241637693 seconds
```

Go

`package main import "fmt" type pair struct{ x, y int } func main() {	//const max = 100	// Use 1685 (the highest with a unique answer) instead	// of 100 just to make it work a little harder :).	const max = 1685	var all []pair	for a := 2; a < max; a++ {		for b := a + 1; b < max-a; b++ {			all = append(all, pair{a, b})		}	}	fmt.Println("There are", len(all), "pairs where a+b <", max, "(and a<b)")	products := countProducts(all) 	// Those for which no sum decomposition has unique product to are	// S mathimatician's possible pairs.	var sPairs []pairpairs:	for _, p := range all {		s := p.x + p.y		// foreach a+b=s (a<b)		for a := 2; a < s/2+s&1; a++ {			b := s - a			if products[a*b] == 1 {				// Excluded because P would have a unique product				continue pairs			}		}		sPairs = append(sPairs, p)	}	fmt.Println("S starts with", len(sPairs), "possible pairs.")	//fmt.Println("S pairs:", sPairs)	sProducts := countProducts(sPairs) 	// Look in sPairs for those with a unique product to get	// P mathimatician's possible pairs.	var pPairs []pair	for _, p := range sPairs {		if sProducts[p.x*p.y] == 1 {			pPairs = append(pPairs, p)		}	}	fmt.Println("P then has", len(pPairs), "possible pairs.")	//fmt.Println("P pairs:", pPairs)	pSums := countSums(pPairs) 	// Finally, look in pPairs for those with a unique sum	var final []pair	for _, p := range pPairs {		if pSums[p.x+p.y] == 1 {			final = append(final, p)		}	} 	// Nicely show any answers.	switch len(final) {	case 1:		fmt.Println("Answer:", final[0].x, "and", final[0].y)	case 0:		fmt.Println("No possible answer.")	default:		fmt.Println(len(final), "possible answers:", final)	}} func countProducts(list []pair) map[int]int {	m := make(map[int]int)	for _, p := range list {		m[p.x*p.y]++	}	return m} func countSums(list []pair) map[int]int {	m := make(map[int]int)	for _, p := range list {		m[p.x+p.y]++	}	return m} // not used, manually inlined abovefunc decomposeSum(s int) []pair {	pairs := make([]pair, 0, s/2)	for a := 2; a < s/2+s&1; a++ {		pairs = append(pairs, pair{a, s - a})	}	return pairs}`
Output:

For x + y < 100 (`max = 100`):

```There are 2304 pairs where a+b < 100 (and a<b)
S starts with 145 possible pairs.
P then has 86 possible pairs.
```

For x + y < 1685 (`max = 1685`):

```There are 706440 pairs where a+b < 1685 (and a<b)
S starts with 50485 possible pairs.
P then has 17485 possible pairs.
```

Run-time ~1 msec and ~600 msec respectively. Could be slightly faster if the slices and maps were given an estimated capacity to start (e.g. (max/2)² for all pairs) to avoid re-allocations (and resulting copies).

Translation of: D
`import Data.List (intersect) s1, s2, s3, s4 :: [(Int, Int)]s1 = [(x, y) | x <- [1 .. 100], y <- [1 .. 100], 1 < x && x < y && x + y < 100] add, mul :: (Int, Int) -> Intadd (x, y) = x + ymul (x, y) = x * y sumEq, mulEq :: (Int, Int) -> [(Int, Int)]sumEq p = filter (\q -> add q == add p) s1mulEq p = filter (\q -> mul q == mul p) s1 s2 = filter (\p -> all (\q -> (length \$ mulEq q) /= 1) (sumEq p)) s1s3 = filter (\p -> length (mulEq p `intersect` s2) == 1) s2s4 = filter (\p -> length (sumEq p `intersect` s3) == 1) s3 main = print s4`
Output:
`[(4,13)]`

Or, to illustrate some of the available variants, it turns out that we can double performance by slightly rearranging the filters in sumEq and mulEq. It also proves fractionally faster to shed some of the of outer list comprehension sugaring, using >>= or concatMap directly.

For a further doubling of performance, we can redefine add and mul as uncurried versions of (+) and (*).

The y > x condition can usefully be moved upstream – dropping it from the test, and redefining the range of y as [x + 1 .. 100] from the start. (The 1 < x test can also be moved out of the test and into the initial generator).

Finally, as we expect and need only one solution, Haskell's lazy evaluation strategy will avoid wasted tests if we request only the first item from the possible solution stream.

`import Data.List (intersect) ------------------ SUM AND PRODUCT PUZZLE ---------------- s1, s2, s3, s4 :: [(Int, Int)]s1 =  [2 .. 100]    >>= \x ->      [succ x .. 100]        >>= \y ->          [ (x, y)            | x + y < 100          ] s2 = filter (all ((1 /=) . length . mulEq) . sumEq) s1 s3 = filter ((1 ==) . length . (`intersect` s2) . mulEq) s2 s4 = filter ((1 ==) . length . (`intersect` s3) . sumEq) s3 sumEq, mulEq :: (Int, Int) -> [(Int, Int)]sumEq p = filter ((add p ==) . add) s1mulEq p = filter ((mul p ==) . mul) s1 add, mul :: (Int, Int) -> Intadd = uncurry (+)mul = uncurry (*) --------------------------- TEST -------------------------main :: IO ()main = print \$ take 1 s4`
Output:
`[(4,13)]`

Java

`package org.rosettacode; import java.util.ArrayList;import java.util.List;  /** * This program applies the logic in the Sum and Product Puzzle for the value  * provided by systematically applying each requirement to all number pairs in  * range. Note that the requirements: (x, y different), (x < y), and  * (x, y > MIN_VALUE) are baked into the loops in run(), sumAddends(), and  * productFactors(), so do not need a separate test. Also note that to test a * solution to this logic puzzle, it is suggested to test the condition with * maxSum = 1685 to ensure that both the original solution (4, 13) and the * additional solution (4, 61), and only these solutions, are found. Note * also that at 1684 only the original solution should be found! */public class SumAndProductPuzzle {    private final long beginning;    private final int maxSum;    private static final int MIN_VALUE = 2;    private List<int[]> firstConditionExcludes = new ArrayList<>();    private List<int[]> secondConditionExcludes = new ArrayList<>();     public static void main(String... args){         if (args.length == 0){            new SumAndProductPuzzle(100).run();            new SumAndProductPuzzle(1684).run();            new SumAndProductPuzzle(1685).run();        } else {            for (String arg : args){                try{                    new SumAndProductPuzzle(Integer.valueOf(arg)).run();                } catch (NumberFormatException e){                    System.out.println("Please provide only integer arguments. " +                            "Provided argument " + arg + " was not an integer. " +                            "Alternatively, calling the program with no arguments " +                            "will run the puzzle where maximum sum equals 100, 1684, and 1865.");                }            }        }    }     public SumAndProductPuzzle(int maxSum){        this.beginning = System.currentTimeMillis();        this.maxSum = maxSum;        System.out.println("Run with maximum sum of " + String.valueOf(maxSum) +                 " started at " + String.valueOf(beginning) + ".");    }     public void run(){        for (int x = MIN_VALUE; x < maxSum - MIN_VALUE; x++){            for (int y = x + 1; y < maxSum - MIN_VALUE; y++){                 if (isSumNoGreaterThanMax(x,y) &&                    isSKnowsPCannotKnow(x,y) &&                    isPKnowsNow(x,y) &&                    isSKnowsNow(x,y)                    ){                    System.out.println("Found solution x is " + String.valueOf(x) + " y is " + String.valueOf(y) +                             " in " + String.valueOf(System.currentTimeMillis() - beginning) + "ms.");                }            }        }        System.out.println("Run with maximum sum of " + String.valueOf(maxSum) +                 " ended in " + String.valueOf(System.currentTimeMillis() - beginning) + "ms.");    }     public boolean isSumNoGreaterThanMax(int x, int y){        return x + y <= maxSum;    }     public boolean isSKnowsPCannotKnow(int x, int y){         if (firstConditionExcludes.contains(new int[] {x, y})){            return false;        }         for (int[] addends : sumAddends(x, y)){            if ( !(productFactors(addends[0], addends[1]).size() > 1) ) {                firstConditionExcludes.add(new int[] {x, y});                return false;            }        }        return true;    }     public boolean isPKnowsNow(int x, int y){         if (secondConditionExcludes.contains(new int[] {x, y})){            return false;        }         int countSolutions = 0;        for (int[] factors : productFactors(x, y)){            if (isSKnowsPCannotKnow(factors[0], factors[1])){                countSolutions++;            }        }         if (countSolutions == 1){            return true;        } else {            secondConditionExcludes.add(new int[] {x, y});            return false;        }    }     public boolean isSKnowsNow(int x, int y){         int countSolutions = 0;        for (int[] addends : sumAddends(x, y)){            if (isPKnowsNow(addends[0], addends[1])){                countSolutions++;            }        }        return countSolutions == 1;    }     public List<int[]> sumAddends(int x, int y){         List<int[]> list = new ArrayList<>();        int sum = x + y;         for (int addend = MIN_VALUE; addend < sum - addend; addend++){            if (isSumNoGreaterThanMax(addend, sum - addend)){                list.add(new int[]{addend, sum - addend});            }        }        return list;    }     public List<int[]> productFactors(int x, int y){         List<int[]> list = new ArrayList<>();        int product = x * y;         for (int factor = MIN_VALUE; factor < product / factor; factor++){            if (product % factor == 0){                if (isSumNoGreaterThanMax(factor, product / factor)){                    list.add(new int[]{factor, product / factor});                }            }        }        return list;    }}`
Output:
```Run with maximum sum of 100 started at 1492436207694.
Found solution x is 4 y is 13 in 7ms.
Run with maximum sum of 100 ended in 54ms.
Run with maximum sum of 1684 started at 1492436207748.
Found solution x is 4 y is 13 in 9084ms.
Run with maximum sum of 1684 ended in 8234622ms.
Run with maximum sum of 1685 started at 1492444442371.
Found solution x is 4 y is 13 in 8922ms.
Found solution x is 4 y is 61 in 8939ms.
Run with maximum sum of 1685 ended in 8013991ms.```

JavaScript

ES5

`(function () {    'use strict';     // GENERIC FUNCTIONS     // concatMap :: (a -> [b]) -> [a] -> [b]    var concatMap = function concatMap(f, xs) {            return [].concat.apply([], xs.map(f));        },         // curry :: ((a, b) -> c) -> a -> b -> c        curry = function curry(f) {            return function (a) {                return function (b) {                    return f(a, b);                };            };        },         // intersectBy :: (a - > a - > Bool) - > [a] - > [a] - > [a]        intersectBy = function intersectBy(eq, xs, ys) {            return xs.length && ys.length ? xs.filter(function (x) {                return ys.some(curry(eq)(x));            }) : [];        },         // range :: Int -> Int -> Maybe Int -> [Int]        range = function range(m, n, step) {            var d = (step || 1) * (n >= m ? 1 : -1);            return Array.from({                length: Math.floor((n - m) / d) + 1            }, function (_, i) {                return m + i * d;            });        };     // PROBLEM FUNCTIONS     // add, mul :: (Int, Int) -> Int    var add = function add(xy) {            return xy[0] + xy[1];        },        mul = function mul(xy) {            return xy[0] * xy[1];        };     // sumEq, mulEq :: (Int, Int) -> [(Int, Int)]    var sumEq = function sumEq(p) {            var addP = add(p);            return s1.filter(function (q) {                return add(q) === addP;            });        },        mulEq = function mulEq(p) {            var mulP = mul(p);            return s1.filter(function (q) {                return mul(q) === mulP;            });        };     // pairEQ :: ((a, a) -> (a, a)) -> Bool    var pairEQ = function pairEQ(a, b) {        return a[0] === b[0] && a[1] === b[1];    };     // MAIN     // xs :: [Int]    var xs = range(1, 100);     // s1 s2, s3, s4 :: [(Int, Int)]    var s1 = concatMap(function (x) {            return concatMap(function (y) {                return 1 < x && x < y && x + y < 100 ? [                    [x, y]                ] : [];            }, xs);        }, xs),         s2 = s1.filter(function (p) {            return sumEq(p).every(function (q) {                return mulEq(q).length > 1;            });        }),         s3 = s2.filter(function (p) {            return intersectBy(pairEQ, mulEq(p), s2).length === 1;        }),         s4 = s3.filter(function (p) {            return intersectBy(pairEQ, sumEq(p), s3).length === 1;        });     return s4;})(); `
Output:
`[[4, 13]]`

(Finished in 0.69s)

ES6

`(() => {    "use strict";     // ------------- SUM AND PRODUCT PUZZLE --------------     // main :: IO ()    const main = () => {        const            // xs :: [Int]            xs = enumFromTo(1)(100),             // s1 s2, s3, s4 :: [(Int, Int)]            s1 = xs.flatMap(                x => xs.flatMap(y =>                    (1 < x) && (x < y) && 100 > (x + y) ? [                        [x, y]                    ] : []                )            ),            s2 = s1.filter(                p => sumEq(p, s1).every(                    q => 1 < mulEq(q, s1).length                )            ),            s3 = s2.filter(                p => 1 === intersectBy(pairEQ)(                    mulEq(p, s1)                )(s2).length            );         return s3.filter(            p => 1 === intersectBy(pairEQ)(                sumEq(p, s1)            )(s3).length        );    };     // ---------------- PROBLEM FUNCTIONS ----------------     // add, mul :: (Int, Int) -> Int    const        add = xy => xy[0] + xy[1],        mul = xy => xy[0] * xy[1],         // sumEq, mulEq :: (Int, Int) ->        // [(Int, Int)] -> [(Int, Int)]        sumEq = (p, s) => {            const addP = add(p);             return s.filter(q => add(q) === addP);        },        mulEq = (p, s) => {            const mulP = mul(p);             return s.filter(q => mul(q) === mulP);        },         // pairEQ :: ((a, a) -> (a, a)) -> Bool        pairEQ = a => b => (            a[0] === b[0]        ) && (a[1] === b[1]);      // ---------------- GENERIC FUNCTIONS ----------------     // enumFromTo :: Int -> Int -> [Int]    const enumFromTo = m =>        n => Array.from({            length: 1 + n - m        }, (_, i) => m + i);      // intersectBy :: (a -> a -> Bool) -> [a] -> [a] -> [a]    const intersectBy = eqFn =>        // The intersection of the lists xs and ys        // in terms of the equality defined by eq.        xs => ys => xs.filter(            x => ys.some(eqFn(x))        );     // MAIN ---    return main();})();`
Output:
`[[4, 13]]`

jq

Works with: jq

Works with: gojq (the Go implementation of jq)

A transcription from the problem statement, with these helper functions:

` # For readability:def collect(c): map(select(c)); # stream-oriented checks:def hasMoreThanOne(s): [limit(2;s)] | length > 1; def hasOne(s): [limit(2;s)] | length == 1; def prod: .[0] * .[1]; ## A stream of admissible [x,y] valuesdef xy:  [range(2;50) as \$x  # 1 < X < Y < 100   | range(\$x+1; 101-\$x) as \$y   | [\$x, \$y] ] ; # The stream of [x,y] pairs matching "S knows the sum is \$sum"def sumEq(\$sum): select( \$sum == add ); # The stream of [x,y] pairs matching "P knows the product is \$prod"def prodEq(\$p): select( \$p == prod ); ## The solver:def solve:  xy as \$s0 # S says P does not know:  | \$s0  | collect(add as \$sum      | all( \$s0[]|sumEq(\$sum);             prod as \$p             | hasMoreThanOne(\$s0[] | prodEq(\$p)))) as \$s1 # P says: Now I know:  | \$s1  | collect(prod as \$prod | hasOne( \$s1[]|prodEq(\$prod)) ) as \$s2 # S says: Now I also know  | \$s2[]  | select(add as \$sum | hasOne( \$s2[] | sumEq(\$sum)) ) ; solve `
Output:
`[4,13]`

Julia

From the awk/sidef version. It is also possible to use filters as in the Scala solution, but although less verbose, using filters would be much slower in Julia, which often favors fast for loops over lists for speed.

` using Primes function satisfy1(x::Integer)    prmslt100 = primes(100)    for i in 2:(x ÷ 2)        if i ∈ prmslt100 && x - i ∈ prmslt100            return false        end    end    return trueend function satisfy2(x::Integer)    once = false    for i in 2:isqrt(x)        if x % i == 0            j = x ÷ i            if 2 < j < 100 && satisfy1(i + j)                if once return false end                once = true            end        end    end    return onceend function satisfyboth(x::Integer)    if !satisfy1(x) return 0 end    found = 0    for i in 2:(x ÷ 2)        if satisfy2(i * (x - i))            if found > 0 return 0 end            found = i        end    end    return foundend for i in 2:99    if (j = satisfyboth(i)) > 0        println("Solution: (\$j, \$(i - j))")    endend`
Output:
`Solution: (4, 13)`

Kotlin

`// version 1.1.4-3 data class P(val x: Int, val y: Int, val sum: Int, val prod: Int) fun main(args: Array<String>) {    val candidates = mutableListOf<P>()        for (x in 2..49) {        for (y in x + 1..100 - x) {            candidates.add(P(x, y, x + y, x * y))        }    }     val sums = candidates.groupBy { it.sum }    val prods = candidates.groupBy { it.prod }     val fact1 = candidates.filter { sums[it.sum]!!.all { prods[it.prod]!!.size > 1 } }    val fact2 = fact1.filter { prods[it.prod]!!.intersect(fact1).size == 1 }    val fact3 = fact2.filter { sums[it.sum]!!.intersect(fact2).size == 1 }    print("The only solution is : ")    for ((x, y, _, _) in fact3) println("x = \$x, y = \$y")    }`
Output:
```The only solution is : x = 4, y = 13
```

Lua

Translation of: C++
`function print_count(t)    local cnt = 0    for k,v in pairs(t) do        cnt = cnt + 1    end    print(cnt .. ' candidates')end function make_pair(a,b)    local t = {}    table.insert(t, a) -- 1    table.insert(t, b) -- 2    return tend function setup()    local candidates = {}    for x = 2, 98 do        for y = x + 1, 98 do            if x + y <= 100 then                local p = make_pair(x, y)                table.insert(candidates, p)            end        end    end    return candidatesend function remove_by_sum(candidates, sum)    for k,v in pairs(candidates) do        local s = v[1] + v[2]        if s == sum then            table.remove(candidates, k)        end    endend function remove_by_prod(candidates, prod)    for k,v in pairs(candidates) do        local p = v[1] * v[2]        if p == prod then            table.remove(candidates, k)        end    endend function statement1(candidates)    local unique = {}    for k,v in pairs(candidates) do        local prod = v[1] * v[2]        if unique[prod] ~= nil then            unique[prod] = unique[prod] + 1        else            unique[prod] = 1        end    end     local done    repeat        done = true        for k,v in pairs(candidates) do            local prod = v[1] * v[2]            if unique[prod] == 1 then                local sum = v[1] + v[2]                remove_by_sum(candidates, sum)                done = false                break            end        end    until doneend function statement2(candidates)    local unique = {}    for k,v in pairs(candidates) do        local prod = v[1] * v[2]        if unique[prod] ~= nil then            unique[prod] = unique[prod] + 1        else            unique[prod] = 1        end    end     local done    repeat        done = true        for k,v in pairs(candidates) do            local prod = v[1] * v[2]            if unique[prod] > 1 then                remove_by_prod(candidates, prod)                done = false                break            end        end    until doneend function statement3(candidates)    local unique = {}    for k,v in pairs(candidates) do        local sum = v[1] + v[2]        if unique[sum] ~= nil then            unique[sum] = unique[sum] + 1        else            unique[sum] = 1        end    end     local done    repeat        done = true        for k,v in pairs(candidates) do            local sum = v[1] + v[2]            if unique[sum] > 1 then                remove_by_sum(candidates, sum)                done = false                break            end        end    until doneend function main()    local candidates = setup()    print_count(candidates)     statement1(candidates)    print_count(candidates)     statement2(candidates)    print_count(candidates)     statement3(candidates)    print_count(candidates)     for k,v in pairs(candidates) do        local sum = v[1] + v[2]        local prod = v[1] * v[2]        print("a=" .. v[1] .. ", b=" .. v[2] .. "; S=" .. sum .. ", P=" .. prod)    endend main()`
Output:
```2352 candidates
145 candidates
86 candidates
1 candidates
a=4, b=13; S=17, P=52```

Nim

`import sequtils, sets, sugar, tables var  xycandidates = toSeq(2..98)  sums = collect(initHashSet, for s in 5..100: {s})   # Set of possible sums.  factors: Table[int, seq[(int, int)]]                # Mapping product -> list of factors. # Build the factor mapping.for i in 0..<xycandidates.high:  let x = xycandidates[i]  for j in (i + 1)..xycandidates.high:    let y = xycandidates[j]    factors.mgetOrPut(x * y, @[]).add (x, y) iterator terms(n: int): (int, int) =  ## Yield the possible terms (x, y) of a given sum.  for x in 2..(n - 1) div 2:    yield (x, n - x) # S says "P does not know X and Y."# => For every decomposition of S, there is no product with a single decomposition.for s in toSeq(sums):  for (x, y) in s.terms():    let p = x * y    if factors[p].len == 1:      sums.excl s      break # P says "Now I know X and Y."# => P has only one decomposition with sum in "sums".for p in toSeq(factors.keys):  var sums = collect(initHashSet):               for (x, y) in factors[p]:                 if x + y in sums: {x + y}  if card(sums) > 1: factors.del p # S says "Now I also know X and Y."# => S has only one decomposition with product in "factors".for s in toSeq(sums):  var prods = collect(initHashSet):                for (x, y) in s.terms():                  if x * y in factors: {x * y}  if card(prods) > 1: sums.excl s # Now, combine the sums and the products.for s in sums:  for (x, y) in s.terms:    if x * y in factors: echo (x, y)`
Output:
`(4, 13)`

ooRexx

version 1

Translation of: REXX

for comments see REXX version 4.

`all      =.set~newCall time 'R'cnt.=0do a=2 to 100  do b=a+1 to 100-2    p=a b    if a+b>100 then leave b    all~put(p)    prd=a*b    cnt.prd+=1    End  EndSay "There are" all~items "pairs where X+Y <=" max "(and X<Y)" spairs=.set~newDo Until all~items=0  do p over all    d=decompositions(p)    If take Then      spairs=spairs~union(d)    dif=all~difference(d)    Leave    End  all=dif  endSay "S starts with" spairs~items "possible pairs." sProducts.=0Do p over sPairs  Parse Var p x y  prod=x*y  sProducts.prod+=1  End pPairs=.set~newDo p over sPairs  Parse Var p xb yb  prod=xb*yb  If sProducts.prod=1 Then    pPairs~put(p)  EndSay "P then has" pPairs~items "possible pairs." Sums.=0Do p over pPairs  Parse Var p xc yc  sum=xc+yc  Sums.sum+=1  End final=.set~newDo p over pPairs  Parse Var p x y  sum=x+y  If Sums.sum=1 Then    final~put(p)  End si=0Do p Over final  si+=1  sol.si=p  EndSelect  When final~items=1 Then Say "Answer:" sol.1  When final~items=0 Then Say "No possible answer."  Otherwise Do;            Say final~items "possible answers:"                           Do p over final                             Say p                             End    End  EndSay "Elapsed time:" time('E') "seconds"Exit decompositions: Procedure Expose cnt. take spairs  epairs=.set~new  Use Arg p  Parse Var p aa bb  s=aa+bb  take=1  Do xa=2 To s/2    ya=s-xa    pp=xa ya    epairs~put(pp)    prod=xa*ya    If cnt.prod=1 Then      take=0    End  return epairs`
Output:
```There are 2352 pairs where X+Y <= MAX (and X<Y)
S starts with 145 possible pairs.
P then has 86 possible pairs.
Elapsed time: 0.016000 seconds```

version 2

Uses objects for storing the number pairs. Note the computed hash value and the == method (required to make the set difference work)

`all      =.set~newCall time 'R'cnt.=0do a=2 to 100  do b=a+1 to 100-2    p=.pairs~new(a,b)    if p~sum>100 then leave b    all~put(p)    prd=p~prod    cnt.prd+=1    End  EndSay "There are" all~items "pairs where X+Y <=" max "(and X<Y)" spairs=.set~newDo Until all~items=0  do p over all    d=decompositions(p)    If take Then      spairs=spairs~union(d)    dif=all~difference(d)    Leave    End  all=dif  endSay "S starts with" spairs~items "possible pairs." sProducts.=0Do p over sPairs  prod=p~prod  sProducts.prod+=1  End pPairs=.set~newDo p over sPairs  prod=p~prod  If sProducts.prod=1 Then    pPairs~put(p)  EndSay "P then has" pPairs~items "possible pairs." Sums.=0Do p over pPairs  sum=p~sum  Sums.sum+=1  End final=.set~newDo p over pPairs  sum=p~sum  If Sums.sum=1 Then    final~put(p)  End si=0Do p Over final  si+=1  sol.si=p  EndSelect  When final~items=1 Then Say "Answer:" sol.1~string  When final~items=0 Then Say "No possible answer."  Otherwise Do;            Say final~items "possible answers:"                           Do p over final                             Say p~string                             End    End  EndSay "Elapsed time:" time('E') "seconds"Exit decompositions: Procedure Expose cnt. take spairs  epairs=.set~new  Use Arg p  s=p~sum  take=1  Do xa=2 To s/2    ya=s-xa    pp=.pairs~new(xa,ya)    epairs~put(pp)    prod=pp~prod    If cnt.prod=1 Then      take=0    End  return epairs ::class pairs::attribute a        -- allow access to attribute::attribute b        -- allow access to attribute::attribute sum      -- allow access to attribute::attribute prod     -- allow access to attribute -- only the strict equality form is needed for the collection classes,::method "=="  expose a b  use strict arg other  return a == other~a & b == other~b -- not needed to make the set difference work, but added for completeness::method "\=="  expose a b  use strict arg other  return a \== other~a | b \== other~b ::method hashCode  expose hash  return hash ::method init        -- create pair, calculate sum, product                     -- and index (blank delimited values)  expose hash a b sum prod oid  use arg a, b  hash = a~hashCode~bitxor(b~hashCode) -- create hash value  sum =a+b           -- sum  prod=a*b           -- product ::method string      -- this creates the string to be shown  expose a b  return "[x="||a",y="||b"]"`
Output:
```There are 2352 pairs where X+Y <= MAX (and X<Y)
S starts with 145 possible pairs.
P then has 86 possible pairs.
Elapsed time: 0.079000 seconds```

Perl

Translation of: Raku
`use List::Util qw(none); sub grep_unique {    my(\$by, @list) = @_;    my @seen;    for (@list) {        my \$x = &\$by(@\$_);        \$seen[\$x]= defined \$seen[\$x] ? 0 : join ' ', @\$_;    }    grep { \$_ } @seen;} sub sums {    my(\$n) = @_;    my @sums;    push @sums, [\$_, \$n - \$_] for 2 .. int \$n/2;    @sums;} sub sum     { \$_[0] + \$_[1] }sub product { \$_[0] * \$_[1] } for \$i (2..97) {    push @all_pairs, map { [\$i, \$_] } \$i + 1..98} # Fact 1:%p_unique = map { \$_ => 1 } grep_unique(\&product, @all_pairs);for my \$p (@all_pairs) {    push @s_pairs, [@\$p] if none { \$p_unique{join ' ', @\$_} } sums sum @\$p;} # Fact 2:@p_pairs = map { [split ' ', \$_] } grep_unique(\&product, @s_pairs); # Fact 3:@final_pair = grep_unique(\&sum, @p_pairs); printf "X = %d, Y = %d\n", split ' ', \$final_pair[0];`
Output:
`X = 4, Y = 13`

Phix

Translation of: AWK

Runs in 0.03s

```with javascript_semantics
function satisfies_statement1(integer s)
-- S says: P does not know the two numbers.
-- Given s, for /all/ pairs (a,b), a+b=s, 2<=a,b<=99, at least one of a or b is composite
for a=2 to floor(s/2) do
if is_prime(a) and is_prime(s-a) then
return false
end if
end for
return true
end function

function satisfies_statement2(integer p)
-- P says: Now I know the two numbers.
-- Given p, for /all/ pairs (a,b), a*b=p, 2<=a,b<=99, exactly one pair satisfies statement 1
bool winner = false
for i=2 to floor(sqrt(p)) do
if mod(p,i)=0 then
integer j = floor(p/i)
if 2<=j and j<=99 then
if satisfies_statement1(i+j) then
if winner then return false end if
winner = true
end if
end if
end if
end for
return winner
end function

function satisfies_statement3(integer s)
-- S says: Now I know the two numbers.
-- Given s, for /all/ pairs (a,b), a+b=s, 2<=a,b<=99, exactly one pair satisfies statements 1 and 2
integer winner = 0
if satisfies_statement1(s) then
for a=2 to floor(s/2) do
if satisfies_statement2(a*(s-a)) then
if winner then return 0 end if
winner = a
end if
end for
end if
return winner
end function

for s=2 to 100 do
integer a = satisfies_statement3(s)
if a!=0 then
printf(1,"%d (%d+%d)\n",{s,a,s-a})
end if
end for
```
Output:
```17 (4+13)
```

Picat

` main =>    N = 98,     PD = new_array(N*N), % PD[I] = no. of product decompositions of I    foreach(I in 1..N*N) PD[I] = 0 end,     foreach(X in 2..N-1, Y in X+1..N) PD[X * Y] := PD[X * Y] + 1 end,      % Fact 1: S says "P does not know X and Y.", i.e.    % For every possible sum decomposition of the number X+Y, the product has in turn more than one product decomposition:    Solutions1 = [[X,Y] : X in 2..N-1, Y in X+1..100-X, foreach(XX in 2..X+Y-3) PD[XX * (X+Y-XX)] > 1 end],      % Fact 2: P says "Now I know X and Y.", i.e.    % The number X*Y has only one product decomposition for which fact 1 is true:     Solutions2 = [[X,Y] : [X,Y] in Solutions1, foreach([XX,YY] in Solutions1, XX * YY = X * Y) XX = X, YY = Y end],      % Fact 3: S says "Now I also know X and Y.", i.e.    % The number X+Y has only one sum decomposition for which fact 2 is true.    Solutions3 = [[X,Y] : [X,Y] in Solutions2, foreach([XX,YY] in Solutions2, XX + YY = X + Y) XX = X, YY = Y end],      println(Solutions3). `

Output:

`[[4,13]]`

Python

Based on the Python solution from Wikipedia:

`#!/usr/bin/env python from collections import Counter def decompose_sum(s):    return [(a,s-a) for a in range(2,int(s/2+1))] # Generate all possible pairsall_pairs = set((a,b) for a in range(2,100) for b in range(a+1,100) if a+b<100) # Fact 1 --> Select pairs for which all sum decompositions have non-unique productproduct_counts = Counter(c*d for c,d in all_pairs)unique_products = set((a,b) for a,b in all_pairs if product_counts[a*b]==1)s_pairs = [(a,b) for a,b in all_pairs if    all((x,y) not in unique_products for (x,y) in decompose_sum(a+b))] # Fact 2 --> Select pairs for which the product is uniqueproduct_counts = Counter(c*d for c,d in s_pairs)p_pairs = [(a,b) for a,b in s_pairs if product_counts[a*b]==1] # Fact 3 --> Select pairs for which the sum is uniquesum_counts = Counter(c+d for c,d in p_pairs)final_pairs = [(a,b) for a,b in p_pairs if sum_counts[a+b]==1] print(final_pairs)`
Output:
`[(4, 13)]`

Racket

Translation of: D
To calculate the results faster this program use memorization. So it has a modified version of `sum=` and `mul=` to increase the chances of reusing the results.
`#lang racket(define-syntax-rule (define/mem (name args ...) body ...)  (begin    (define cache (make-hash))    (define (name args ...)      (hash-ref! cache (list args ...) (lambda () body ...))))) (define (sum p) (+ (first p) (second p)))(define (mul p) (* (first p) (second p))) (define (sum= p s) (filter (lambda (q) (= p (sum q))) s))(define (mul= p s) (filter (lambda (q) (= p (mul q))) s)) (define (puzzle tot)  (printf "Max Sum: ~a\n" tot)  (define s1 (for*/list ([x (in-range 2 (add1 tot))]                         [y (in-range (add1 x) (- (add1 tot) x))])               (list x y)))  (printf "Possible pairs: ~a\n" (length s1))   (define/mem (sumEq/all p) (sum= p s1))  (define/mem (mulEq/all p) (mul= p s1))   (define s2 (filter (lambda (p) (andmap (lambda (q)                                           (not (= (length (mulEq/all (mul q))) 1)))                                         (sumEq/all (sum p))))                     s1))  (printf "Initial pairs for S: ~a\n" (length s2))   (define s3 (filter (lambda (p) (= (length (mul= (mul p) s2)) 1))                   s2))  (displayln (length s3))  (printf "Pairs for P: ~a\n" (length s3))   (define s4 (filter (lambda (p) (= (length (sum= (sum p) s3)) 1))                     s3))  (printf "Final pairs for S: ~a\n" (length s4))   (displayln s4)) (puzzle 100)`
Output:
```Max Sum: 100
Possible pairs: 2352
Initial pairs for S: 145
Pairs for P: 86
Final pairs for S: 1
((4 13))```

Raku

(formerly Perl 6)

Translation of: Python
Works with: Rakudo version 2016.07
`sub grep-unique (&by, @list) { @list.classify(&by).values.grep(* == 1).map(*[0]) }sub sums        (\$n)         { (\$_, \$n - \$_ for 2 .. \$n div 2) }sub sum         ([\$x, \$y])   { \$x + \$y }sub product     ([\$x, \$y])   { \$x * \$y } my @all-pairs = (|(\$_ X \$_+1 .. 98) for 2..97); # Fact 1:my %p-unique := Set.new: map ~*, grep-unique &product, @all-pairs;my @s-pairs = @all-pairs.grep: { none (%p-unique{~\$_} for sums sum \$_) }; # Fact 2:my @p-pairs = grep-unique &product, @s-pairs; # Fact 3:my @final-pairs = grep-unique &sum, @p-pairs; printf "X = %d, Y = %d\n", |\$_ for @final-pairs;`
Output:
`X = 4, Y = 13`

REXX

version 1

I tried hard to understand/translate the algorithms shown so far (16 Oct 2016) Unfortunately to no avail (not knowing the semantics of the used languages). Finally I was successful by implementing the rules referred to in Wikipedia http://www.win.tue.nl/~gwoegi/papers/freudenthal1.pdf which had a very clear description.

`debug=0If debug Then Do  oid='sppn.txt'; 'erase' oid  EndCall time 'R'all_pairs=''cnt.=0i=0/* first take all possible pairs 2<=x<y with x+y<=100 *//* and compute the respective sums and products       *//* count the number of times a sum or product occurs  */Do x=2 To 98  Do y=x+1 To 100-x    x=right(x,2,0)    y=right(y,2,0)    all_pairs=all_pairs x'/'y    i=i+1    x.i=x    y.i=y    sum=x+y    prd=x*y    cnt.0s.sum=cnt.0s.sum+1    cnt.0p.prd=cnt.0p.prd+1    End  Endn=i/* now compute the possible pairs for each sum sum_d.sum *//*                                 and product prd_d.prd *//* also the list of possible sums and products suml, prdl*/sum_d.=''prd_d.=''suml=''prdl=''Do i=1 To n  x=x.i  y=y.i  x=right(x,2,0)  y=right(y,2,0)  sum=x+y  prd=x*y  cnt.0s.x.y=cnt.0s.sum  cnt.0p.x.y=cnt.0p.prd  sum_d.sum=sum_d.sum x'/'y  prd_d.prd=prd_d.prd x'/'y  If wordpos(sum,suml)=0 Then suml=suml sum  If wordpos(prd,prdl)=0 Then prdl=prdl prd  EndSay n 'possible pairs'Call o 'SUM'suml=wordsort(suml)prdl=wordsort(prdl)sumlc=sumlsi=0pi=0Do While sumlc>''  Parse Var sumlc sum sumlc  si=si+1  sum.si=sum  si.sum=si  If sum=17 Then sx=si  temp=prdl  Do While temp>''    Parse Var temp prd temp    If si=1 Then Do      pi=pi+1      prd.pi=prd      pi.prd=pi      If prd=52 Then px=pi      End    A.prd.sum='+'    End  Endsin=sipin=piCall o 'SUM'Do si=1 To sin  Call o f5(si) f3(sum.si)  EndCall o 'PRD'Do pi=1 To pin  Call o f5(pi) f6(prd.pi)  Enda.='-'Do pi=1 To pin  prd=prd.pi  Do si=1 To sin    sum=sum.si    Do sj=1 To words(sum_d.sum)      If wordpos(word(sum_d.sum,sj),prd_d.prd)>0 Then        Parse Value word(sum_d.sum,sj) with x '/' y        prde=x*y        sume=x+y        pa=pi.prde        sa=si.sume        a.pa.sa='+'      End    End  EndCall show '1' Do pi=1 To pin  prow=''  cnt=0  Do si=1 To sin    If a.pi.si='+' Then Do      cnt=cnt+1      pj=pi      sj=si      End    End  If cnt=1 Then    a.pj.sj='1'  EndCall show '2' Do si=1 To sin  Do pi=1 To pin    If a.pi.si='1' Then Leave    End  If pi<=pin Then Do    Do pi=1 To pin      If a.pi.si='+' Then        a.pi.si='2'      End    End  EndCall show '3' Do pi=1 To pin  prow=''  Do si=1 To sin    prow=prow||a.pi.si    End  If count('+',prow)>1 Then Do    Do si=1 To sin      If a.pi.si='+' Then        a.pi.si='3'      End    End  EndCall show '4' Do si=1 To sin  scol=''  Do pi=1 To pin    scol=scol||a.pi.si    End  If count('+',scol)>1 Then Do    Do pi=1 To pin      If a.pi.si='+' Then        a.pi.si='4'      End    End  EndCall show '5' sol=0Do pi=1 To pin  Do si=1 To sin    If a.pi.si='+' Then Do      Say sum.si prd.pi      sum=sum.si      prd=prd.pi      sol=sol+1      End    End  EndSay sol 'solution(s)'Say '            possible pairs'Say 'Product='prd prd_d.52Say '    Sum='sum sum_d.17Say 'The only pair in both lists is 04/13.'Say 'Elapsed time:' time('E') 'seconds'Exitshow:If debug Then Do  Call o 'show' arg(1)  Do pi=1 To 60    ol=''    Do si=1 To 60      ol=ol||a.pi.si      End    Call o ol    End  Say 'a.'px'.'sx'='a.px.sx  EndReturn Exito: Return lineout(oid,arg(1))f3: Return format(arg(1),3)f4: Return format(arg(1),4)f5: Return format(arg(1),5)f6: Return format(arg(1),6) count: Procedure  Parse Arg c,s  s=translate(s,c,c||xrange('00'x,'ff'x))  s=space(s,0)  Return length(s)wordsort: Procedure/*********************************************************************** Sort the list of words supplied as argument. Return the sorted list**********************************************************************/  Parse Arg wl  wa.=''  wa.0=0  Do While wl<>''    Parse Var wl w wl    Do i=1 To wa.0      If wa.i>w Then Leave      End    If i<=wa.0 Then Do      Do j=wa.0 To i By -1        ii=j+1        wa.ii=wa.j        End      End    wa.i=w    wa.0=wa.0+1    End  swl=''  Do i=1 To wa.0    swl=swl wa.i    End  /* Say swl */  Return strip(swl)`
Output:
```2352 possible pairs
17 52
1 solution(s)
possible pairs
Product=52  02/26 04/13
Sum=17  02/15 03/14 04/13 05/12 06/11 07/10 08/09
The only pair in both lists is 04/13.
Elapsed time: 4.891000 seconds```

version 2

Translation of: AWK
`Call time 'R'Do s=2 To 100  a=satisfies_statement3(s)  If a>0 Then Do    p=a*(s-a)    Say a'/'||(s-a) 's='s 'p='p    End  EndSay 'Elapsed time:' time('E') 'seconds'Exit satisfies_statement1: Procedure  Parse Arg s  Do a=2 To s/2    If is_prime(a) & is_prime(s-a) Then      Return 0    End  Return 1 satisfies_statement2: Procedure  Parse Arg p  winner=0  Do i=2 By 1 While i**2<p    If p//i=0 Then Do      j=p%i      If 2<=j & j<=99 Then Do        if satisfies_statement1(i+j) Then Do          if winner Then            Return 0          winner=1          End        End      End    End  Return winner satisfies_statement3: Procedure  Parse Arg s  winner=0  If satisfies_statement1(s)=0 Then    Return 0  Do a=2 To s/2    b=s-a    If satisfies_statement2(a*b) Then Do      If winner>0 Then        Return 0      winner=a      End    End  Return winner is_prime: Procedure  call Trace 'O'  Parse Arg x  If x<=3 Then Return 1  i=2  Do i=2 By 1 While i**2<=x    If datatype(x/i,'W') Then Return 0    End  Return 1`
Output:
```4/13 s=17 p=52
Elapsed time: 0.078000 seconds```

version 3

Translation of: GO
`/*---------------------------------------------------------------------* X and Y are two different whole numbers greater than 1.* Their sum is no greater than 100, and Y is greater than X.* S and P are two mathematicians (and consequently perfect logicians);* S knows the sum X+Y and P knows the product X*Y.* Both S and P know all the information in this paragraph.** The following conversation occurs:** * S says "P does not know X and Y."* * P says "Now I know X and Y."* * S says "Now I also know X and Y!"** What are X and Y?*--------------------------------------------------------------------*/Call time 'R'max=100Products.=0all=''Do x=2 To max  Do y=x+1 To max-2    If x+y<=100 Then Do      all=all x'/'y      prod=x*y; Products.prod=Products.prod+1      End    End  EndSay "There are" words(all) "pairs where X+Y <=" max "(and X<Y)"/*---------------------------------------------------------------------* First eliminate all pairs where the product is unique:* For each pair we look at the decompositions of the sum (x+y).* If for any of these decompositions (xa/ya) the product is unique* then x/y cannot be the solution of the puzzle and we eliminate it* from the list of possible pairs*--------------------------------------------------------------------*/sPairs=''Do i=1 To words(all)  xy=word(all,i)  Parse Var xy x '/' Y  Parse Var xy xx '/' Yy  s=x+y  take=1  Do xa=2 To s/2    ya=s-xa    prod=xa*ya    If products.prod=1 Then Do      take=0      Iterate i      End    End  If take Then    sPairs=sPairs xy  EndSay "S starts with" words(sPairs) "possible pairs." /*---------------------------------------------------------------------* From the REMAINING pairs take only these where the product is unique:* For each pair we look at the decompositions of the known product.* If for any of these decompositions (xb/yb) the product is unique* then xb/yb can be the solution of the puzzle and we add it* to the list of possible pairs.*--------------------------------------------------------------------*/sProducts.=0Do i=1 To words(sPairs)  xy=word(sPairs,i)  Parse Var xy x '/' y  prod=x*y  sProducts.prod=sProducts.prod+1  EndpPairs=''Do i=1 To words(sPairs)  xy=word(sPairs,i)  Parse Var xy x '/' y  prod=x*y  If sProducts.prod=1 Then    pPairs=pPairs xy  EndSay "P then has" words(pPairs) "possible pairs." /*---------------------------------------------------------------------* From the now REMAINING pairs take only these where the sum is unique* Now we look at all possible pairs and find the one (xc/yc)* with a unique sum which must be the sum we knew from the beginning.* The pair xc/yc is then the solution*--------------------------------------------------------------------*/Sums.=0Do i=1 To words(pPairs)  xy=word(pPairs,i)  Parse Var xy x '/' y  sum=x+y  Sums.sum=Sums.sum+1  Endfinal=''Do i=1 To words(pPairs)  xy=word(pPairs,i)  Parse Var xy x '/' y  sum=x+y  If Sums.sum=1 Then    final = final xy  EndSelect  When words(final)=1 Then Say "Answer:" strip(final)  When words(final)=0 Then Say "No possible answer."  Otherwise Do;            Say words(final) "possible answers:"                           Say strip(final)    End  EndSay "Elapsed time:" time('E') "seconds"Exit`
Output:
```There are 2352 pairs where X+Y <= 100 (and X<Y)
S starts with 145 possible pairs.
P then has 86 possible pairs.
Elapsed time: 0.045000 seconds```

version 4

Now that I have understood the logic (I am neither S nor P) I have created an alternative to version 3.

`/*---------------------------------------------------------------------* X and Y are two different whole numbers greater than 1.* Their sum is no greater than 100, and Y is greater than X.* S and P are two mathematicians (and consequently perfect logicians);* S knows the sum X+Y and P knows the product X*Y.* Both S and P know all the information in this paragraph.** The following conversation occurs:** * S says "P does not know X and Y."* * P says "Now I know X and Y."* * S says "Now I also know X and Y!"** What are X and Y?*--------------------------------------------------------------------*/Call time 'R'max=100Products.=0all=''Do x=2 To max  Do y=x+1 To max-2    If x+y<=100 Then Do      all=all x'/'y      prod=x*y; Products.prod=Products.prod+1      End    End  EndSay "There are" words(all) "pairs where X+Y <=" max "(and X<Y)"/*---------------------------------------------------------------------* First eliminate all pairs where the product is unique:* For each pair we look at the decompositions of the sum (x+y).* If for any of these decompositions (xa/ya) the product is unique* then the given sum cannot be the sum of the pair we are looking for* Otherwise all pairs in the sum's decompositions are eligible.*--------------------------------------------------------------------*/sPairs=''done.=0Do i=1 To words(all)  xy=word(all,i)  If done.xy Then Iterate  Parse Var xy x '/' y  s=x+y  take=1  el=''  Do xa=2 To s/2    ya=s-xa    m=xa'/'ya    done.m=1    el=el m    prod=xa*ya    If products.prod=1 Then      take=0    End  If take Then    sPairs=sPairs el  EndSay "S starts with" words(sPairs) "possible pairs." /*---------------------------------------------------------------------* From the REMAINING pairs take only these where the product is unique:* For each pair we look at the decompositions of the known product.* If for any of these decompositions (xb/yb) the product is unique* then xb/yb can be the solution of the puzzle and we add it* to the list of possible pairs.*--------------------------------------------------------------------*/sProducts.=0Do i=1 To words(sPairs)  xy=word(sPairs,i)  Parse Var xy x '/' y  prod=x*y  sProducts.prod=sProducts.prod+1  EndpPairs=''Do i=1 To words(sPairs)  xy=word(sPairs,i)  Parse Var xy xb '/' yb  prod=xb*yb  If sProducts.prod=1 Then    pPairs=pPairs xy  EndSay "P then has" words(pPairs) "possible pairs." /*---------------------------------------------------------------------* From the now REMAINING pairs take only these where the sum is unique* Now we look at all possible pairs and find the one (xc/yc)* with a unique sum which must be the sum we knew from the beginning.* The pair xc/yc is then the solution*--------------------------------------------------------------------*/Sums.=0Do i=1 To words(pPairs)  xy=word(pPairs,i)  Parse Var xy xc '/' yc  sum=xc+yc  Sums.sum=Sums.sum+1  Endfinal=''Do i=1 To words(pPairs)  xy=word(pPairs,i)  Parse Var xy x '/' y  sum=x+y  If Sums.sum=1 Then    final = final xy  EndSelect  When words(final)=1 Then Say "Answer:" strip(final)  When words(final)=0 Then Say "No possible answer."  Otherwise Do;            Say words(final) "possible answers:"                           Say strip(final)    End  EndSay "Elapsed time:" time('E') "seconds"Exit`
Output:
```There are 2352 pairs where X+Y <= 100 (and X<Y)
S starts with 145 possible pairs.
P then has 86 possible pairs.
Elapsed time: 0.032000 seconds```

version 5, fast

This REXX version is over   ten   times faster than the previous REXX version.

`/*REXX program solves the  Sum and Product Puzzle (also known as the Impossible Puzzle).*/@.= 0;                        h= 100;  @.3= 1    /*assign array default;  assign high P.*/             do j=5  by 2  to h                  /*find all odd primes  ≤  1st argument.*/               do k=3  while k*k<=j;   if j//k==0  then iterate j          /*J ÷ by K ? */               end  /*k*/;             @.j= 1    /*found a net prime number: J          */             end    /*j*/@.2=1                                            /*assign the even prime, ex post facto.*/     do s=2  for h-1;  if C1(s)==0  then iterate /*find and display the puzzle solution.*/     \$= 0;                do m=2  for  s%2 -1    /* [↓]  check for uniqueness of product*/                          if C2(m * (s-m))  then do;  if \$>0  then iterate s;  \$= m;   end                          end   /*m*/     if \$>0  then say  'The numbers are:  '         \$            " and "           s-\$     end   /*s*/if \$==0  then     say  'No solution found.'exit 0                                           /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/C1: procedure expose @.;        parse arg s      /*validate the first puzzle condition. */      do a=2  for s%2-1;        if @.a  then do;   _= s - a;   if @._  then return 0;  end      end;  /*a*/;              return 1/*──────────────────────────────────────────────────────────────────────────────────────*/C2: procedure expose @. h;  parse arg p;    \$= 0 /*validate the second puzzle condition.*/      do j=2  while j*j<p                        /*perform up to the square root of  P. */      if p//j==0  then do;               q= p % j                       if q>=2  then  if q<=h  then  if C1(j+q)  then  if \$  then return 0                                                                             else \$= 1                       end      end   /*j*/;              return \$`
output   when using the default internal input:
```The numbers are:   4  and  13
```

Ruby

Translation of: D
`def add(x,y) x + y enddef mul(x,y) x * y end def sumEq(s,p) s.select{|q| add(*p) == add(*q)} enddef mulEq(s,p) s.select{|q| mul(*p) == mul(*q)} end s1 = (a = *2...100).product(a).select{|x,y| x<y && x+y<100}s2 = s1.select{|p| sumEq(s1,p).all?{|q| mulEq(s1,q).size != 1} }s3 = s2.select{|p| (mulEq(s1,p) & s2).size == 1}p    s3.select{|p| (sumEq(s1,p) & s3).size == 1}`
Output:
```[[4, 13]]
```

Scala

`object ImpossiblePuzzle extends App {  type XY = (Int, Int)  val step0 = for {    x <- 1 to 100    y <- 1 to 100    if 1 < x && x < y && x + y < 100  } yield (x, y)   def sum(xy: XY) = xy._1 + xy._2  def prod(xy: XY) = xy._1 * xy._2  def sumEq(xy: XY) = step0 filter { sum(_) == sum(xy) }  def prodEq(xy: XY) = step0 filter { prod(_) == prod(xy) }   val step2 = step0 filter { sumEq(_) forall { prodEq(_).size != 1 }}  val step3 = step2 filter { prodEq(_).intersect(step2).size == 1 }  val step4 = step3 filter { sumEq(_).intersect(step3).size == 1 }  println(step4)}`
Output:
`Vector((4,13))`

Scheme

` (import (scheme base)        (scheme cxr)        (scheme write)        (srfi 1)) ;; utility method to find unique sum/product in given list(define (unique-items lst key)  (let ((all-items (map key lst)))    (filter (lambda (i) (= 1 (count (lambda (p) (= p (key i)))                                    all-items)))            lst))) ;; list of all (x y x+y x*y) combinations with y > x(define *xy-pairs*   (apply append         (map (lambda (i)                (map (lambda (j)                       (list i j (+ i j) (* i j)))                     (iota (- 98 i) (+ 1 i))))              (iota 96 2)))) ;; S says "P does not know X and Y"(define *products* ; get products which have multiple decompositions  (let ((all-products (map fourth *xy-pairs*)))    (filter (lambda (p) (> (count (lambda (i) (= i p)) all-products) 1))            all-products))) (define *fact-1* ; every x+y has x*y in *products*  (filter (lambda (i)             (every (lambda (p) (memq (fourth p) *products*))                   (filter (lambda (p) (= (third i) (third p))) *xy-pairs*)))          *xy-pairs*)) ;; P says "Now I know X and Y"(define *fact-2* ; find the unique X*Y  (unique-items *fact-1* fourth)) ;; S says "Now I also know X and Y"(define *fact-3* ; find the unique X+Y  (unique-items *fact-2* third)) (display (string-append "Initial pairs: " (number->string (length *xy-pairs*)) "\n"))(display (string-append "After S: " (number->string (length *fact-1*)) "\n"))(display (string-append "After P: " (number->string (length *fact-2*)) "\n"))(display (string-append "After S: " (number->string (length *fact-3*)) "\n"))(display (string-append "X: "                         (number->string (caar *fact-3*))                        " Y: "                        (number->string (cadar *fact-3*))                        "\n")) `
Output:
```Initial pairs: 4656
After S: 145
After P: 86
After S: 1
X: 4 Y: 13
```

Sidef

Translation of: Raku
`func grep_uniq(a, by) { a.group_by{ .(by) }.values.grep{.len == 1}.map{_[0]} }func sums     (n)     { 2 .. n//2 -> map {|i| [i, n-i] } } var pairs = (2..97 -> map {|i| ([i] ~X (i+1 .. 98))... }) var p_uniq = Hash()p_uniq{grep_uniq(pairs, :prod).map { .to_s }...} = () var s_pairs = pairs.grep {|p| sums(p.sum).all { !p_uniq.contains(.to_s) } }var p_pairs = grep_uniq(s_pairs, :prod)var f_pairs = grep_uniq(p_pairs, :sum) f_pairs.each { |p| printf("X = %d, Y = %d\n", p...) }`
Output:
```X = 4, Y = 13
```

Wren

Translation of: Kotlin
Library: Wren-dynamic
Library: Wren-seq
`import "/dynamic" for Tupleimport "/seq" for Lst var P = Tuple.create("P", ["x", "y", "sum", "prod"]) var intersect = Fn.new { |l1, l2|    var l3 = (l1.count < l2.count) ? l1 : l2    var l4 = (l3 == l1) ? l2 : l1    var l5 = []    for (e in l3) if (l4.contains(e)) l5.add(e)    return l5} var candidates = []for (x in 2..49) {    for (y in x + 1..100 - x) {        candidates.add(P.new(x, y, x + y, x * y))    }} var sumGroups  = Lst.groups(candidates) { |c| c.sum }var prodGroups = Lst.groups(candidates) { |c| c.prod }var sumMap = {}for (sumGroup in sumGroups) {    sumMap[sumGroup[0]] = sumGroup[1].map { |l| l[0] }.toList}var prodMap = {}for (prodGroup in prodGroups) {    prodMap[prodGroup[0]] = prodGroup[1].map { |l| l[0] }.toList}var fact1 = candidates.where { |c| sumMap[c.sum].all { |c| prodMap[c.prod].count > 1 } }.toListvar fact2 = fact1.where { |c| intersect.call(prodMap[c.prod], fact1).count == 1 }.toListvar fact3 = fact2.where { |c| intersect.call(sumMap[c.sum], fact2).count == 1 }.toListSystem.write("The only solution is : ")for (p in fact3) System.print("x = %(p.x), y = %(p.y)")`
Output:
```The only solution is : x = 4, y = 13
```

zkl

Damn it Jim, I'm a programmer, not a logician. So I translated the python code found in https://qmaurmann.wordpress.com/2013/08/10/sam-and-polly-and-python/ but I don't understand it. It does seem quite a bit more efficient than the Scala code, on par with the Python code.

`mul:=Utils.Helpers.summer.fp1('*,1); //-->list.reduce('*,1), multiply list itemsvar allPairs=[[(a,b); [2..100]; { [a+1..100] },{ a+b<100 }; ROList]]; // 2,304 pairs sxys,pxys:=Dictionary(),Dictionary();  // hashes of allPairs sums and products: 95,1155foreach xy in (allPairs){ sxys.appendV(xy.sum(),xy); pxys.appendV(xy:mul(_),xy) } sOK:= 'wrap(s){ (not sxys[s].filter1('wrap(xy){ pxys[xy:mul(_)].len()<2 })) };pOK:= 'wrap(p){ 1==pxys[p].filter('wrap([(x,y)]){ sOK(x+y) }).len() };sOK2:='wrap(s){ 1==sxys[s].filter('wrap(xy){ pOK(xy:mul(_)) }).len() };allPairs.filter('wrap([(x,y)]){ sOK(x+y) and pOK(x*y) and sOK2(x+y) }).println();`

[[ ]] denotes list comprehension, filter1 returns (and stops at) the first thing that is "true", 'wrap creates a closure so the "wrapped" code/function can see local variables (read only). In a [function] prototype, the "[(x,y)]xy]" notation says xy is a list like thing, assign the parts to x & y (xy is optional), used here to just to do it both ways. The ":" says take the LHS and stuff it into the "_".

Output:
`L(L(4,13))`